Question: $ g(x) = \int_{0}^{x}\dfrac{t^2 - t}{\sqrt{t}}\,dt\,$ $ g\,'(4)\, =$
Answer: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \dfrac{t^2 - t}{\sqrt{t}}$ is continuous on $[0,4]$. Applying the theorem We're given: $ g(x) = \int_{0}^{x}\dfrac{t^2 - t}{\sqrt{t}}\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =\dfrac{x^2 - x}{\sqrt{x}} $ Evaluating $g'(4)$ $ g'(4)= \dfrac{4^2 - 4}{\sqrt{4}} = \dfrac{12}{2} = 6$ The answer: $g'(4)=6$